Exercise 3.9 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E} \times 3.9$
Question $1 .$
Determine the quadratic equations, whose sum and product of roots are
(i) $-9,20$
(ii) $\frac{5}{3}, 4$
(iii) $\frac{-3}{2},-1$
(iv) $-(2-a)^{2},(a+5)^{2}$
Solution:
If the roots are given, general form of the quadratic equation is $x^{2}-$ (sum of the roots) $x+$ product of the roots $=0$.
(i) Sum of the roots $=-9$
Product of the roots $=20$
The equation $=x^{2}-(-9 x)+20=0$
$\Rightarrow \mathrm{x}^{2}+9 \mathrm{x}+20=0$
(ii) Sum of the roots $=\frac{5}{3}$
Product of the roots $=4$
Required equation $=x^{2}-($ sum of the roots) $x+$ product of the roots
$\begin{aligned}
&=0 \\
&\Rightarrow x^{2}-\frac{5}{3} x+4=0 \\
&\Rightarrow 3 x^{2}-5 x+12=0
\end{aligned}$
(iii) Sum of the roots $=\left(\frac{-3}{2}\right)$
$(\alpha+\beta)=\frac{-3}{2}$
Product of the roots $(\alpha \beta)=(-1)$
Required equation $=x^{2}-(\alpha+\beta) x+\alpha \beta=0$

$\begin{aligned}
&x^{2}-\left(\frac{-3}{2}\right) x-1=0 \\
&2 x^{2}+3 x-2=0 \\
&\text { (iv) } \alpha+\beta=-(2-a)^{2} \\
&\alpha \beta=(a+5)^{2} \\
&\text { Required equation }=x^{2}-(\alpha+\beta) x-\alpha \beta=0 \\
&\Rightarrow x^{2}-\left(-(2-a)^{2}\right) x+(a+5)^{2}=0 \\
&\Rightarrow x^{2}+(2-a)^{2} x+(a+5)^{2}=0
\end{aligned}$

Question $2 .$
Find the sum and product of the roots for each of the following quadratic equations
(i) $x^{2}+3 x-28=0$
(ii) $x^{2}+3 x=0$
(iii) $3+\frac{1}{a}=\frac{10}{a^{2}}$
(iv) $3 y^{2}-y-4=0$
(i) $x^{2}+3 x-28=0$
Answer:
Sum of the roots $(\alpha+\beta)=-3$
Product of the roots $(\alpha \beta)=-28$
(ii) $x^{2}+3 x=0$
Answer:
Sum of the roots $(\alpha+\beta)=-3$
Product of the roots $(\alpha \beta)=0$
(iii) $3+\frac{1}{a}=\frac{10}{a^{2}}$
$3 a^{2}+a=10$
$3 a^{2}+a-10=0$ comparing this with $x^{2}-(\alpha+\beta)$

$x+\alpha \beta=0$
$a^{2}-\left(-\frac{1}{3}\right) a+\left(\frac{-10}{3}\right)=0$
$\alpha+\beta=\frac{-1}{3}$
$\alpha \beta=\frac{-10}{3}$
(iv) $3 y^{2}-y-4=0 \div 3$
$\begin{aligned}
y^{2}-\frac{y}{3}-\frac{4}{3} &=0 \\
y^{2}-\left(\frac{1}{3}\right) y+\left(\frac{-4}{3}\right) &=0 \\
\therefore \alpha+\beta &=\frac{1}{3} \\
\alpha \beta &=\frac{-4}{3}
\end{aligned}$